Problem: $\text C = \left[\begin{array}{rr}1 & 4 \\ 4 & -1 \\ 3 & -2\end{array}\right]$ and $\text D = \left[\begin{array}{rr}-2 & 2 \\ 3 & 0\end{array}\right]$ Let $\text {H = CD}$. Find $\text H$. $ {H = }$
Solution: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{C}$ and the first column of $\text{D}$. $ \text {H}=\left[\begin{array}{rr}{1} & {4} \\ 4 & -1 \\ 3 & -2\end{array}\right]\left[\begin{array}{rr} {-2} & 2 \\ {3} & 0\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(1,4)\cdot(-2,3)\\\\ &=1 \cdot -2 + 4\cdot 3\\\\ &=10 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $1 \cdot 2 + 4\cdot 0 = 2$ (Choice B) B $4 \cdot 2 - 1\cdot 0 = 8$ (Choice C) C $4 \cdot -2 - 1\cdot 3 = -11$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}=\left[\begin{array}{rr}10 & 2 \\ -11 & 8 \\ -12 & 6\end{array}\right] $